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I was doing a question relates to substitution rule under integration.

The question is as follow:

Evaluate $\int{1\over{(1+x^2)^n}}dx, n\in \mathbb{Z}^+$

We have seen that $(\tan^{-1}x)'=\frac{1}{1+x^2}$

Let $x= \tan t, t\in(-\frac{\pi}{2},\frac{\pi}{2})$. Then $\frac{dx}{dt}=\sec^2t$

$$\int\frac{1}{(1+x^2)^n}dx=\int\frac{1}{(1+tan^2t)^n}\sec^2tdt$$ $$=\int\frac{\sec^2t}{\sec^{2n}t}dt=\int\cos^{2n-2}tdt$$

so I got the final answer $\cos^{2n -2} (t) dt$, but the question continues to ask what's the integral when $n=1$.

I substitute $n=1$, and I got $\cos^0 (t)$ which is totally unfamiliar to me.

So what is $\cos^0 (t)$ ?

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    $\begingroup$ Isn't that simply $1$? $\endgroup$ – Arpan Apr 23 '15 at 10:33
  • $\begingroup$ @user33448687 when $n=1$ cant you right away right it as $\arctan (x)$ $\endgroup$ – happymath Apr 23 '15 at 10:34
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$\cos^0(t) = 1$ if $\cos t \ne 0$ and undefined else, so $\int \cos^0 t \ \mathrm dt = \int 1 \ \mathrm dt = t + C$. This results in $$\int \frac1{(1+x^2)^1} \ \mathrm dx = t + C= \arctan x + C$$ wich is consistent with what you know. Note that $\arctan$ is a nicer way of writing $\tan^{-1}$.

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    $\begingroup$ $\arctan$ is also a less misleading way of writing $\tan^{-1}$: The latter borrows the notation $\cdot^{-1}$ used for function inverses, but $\arctan$ is not the inverse of $\tan$ (which is not injective and hence has no inverse) but rather the inverse of the restriction of $\tan$ to $(-\frac{\pi}{2}, \frac{\pi}{2})$. $\endgroup$ – Travis Apr 23 '15 at 11:35
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    $\begingroup$ Domains'll getcha every time. $\endgroup$ – hBy2Py Apr 23 '15 at 12:44
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    $\begingroup$ I don't understand why you write "undefined else": $0^0$ is defined as $1$. And that's important, how would you write power series without this definition? (As in $\exp(x) = \sum_{n=0}^\infty x^n/n!$) $\endgroup$ – Hendrik Vogt May 1 '15 at 4:53
  • $\begingroup$ @HendrikVogt The definition $0^0 = 1$ must be treated with care, because it doesn't work with limits. That's why it usually remains undefined. $\endgroup$ – AlexR May 1 '15 at 7:59
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    $\begingroup$ @AlexR: I'd really like to see a reference to a book where they leave $0^0$ undefined. The reason for my comment was the following: I've been teaching calculus for years, and many students had learned in school that $0^0$ is undefined. Then suddenly we have power series and stuff like that where $0^0$ appears naturally, and one needs to know that it's equal to $1$. So call it my mission to bring the gospel $0^0=1$ to the world. $\endgroup$ – Hendrik Vogt May 1 '15 at 9:55
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For every non-zero number $x$, you have $x^0=1$

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  • $\begingroup$ non-zero complex number, to be exact. $\endgroup$ – user26486 Apr 24 '15 at 0:46

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